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Math 100 105, Fall Term 2010 Midterm Exam November 8th ,2010 Student number: LAST name: First name: Instructions • Do not turn this page over until instructed. You will have 45 minutes for the exam. • You may not use books, notes or electronic devices of any kind. • Solutions should be written clearly, in complete English sentences, showing all your work. • If you are using a result from the textbook, the lectures or the problem sets, state it properly. Signature: 1 2 3 4 Total 1 /18 /6 /6 /10 /40 Math 100 105, Fall 2010 1 Midterm, page 2/7 Problem 1 Short-form answers Show your work and clearly delineate your nal answer. Not all problems are of equal diculty. [3] a. Dierentiate the function y = arctan(x); write your answer as a function of u alone (you may use the formula d tan = 1 + tan2 u). du x If y = arctan(x) then tan(y) = x. Dierentiating with respect to x and using the chain rule we nd y) dy dy dy 1 = d(tan = (1 + tan2 y) dx = (1 + x2 ) dx . Solving for dx we nd dx 1 dy . = dx 1 + x2 [3] b. Find the tangent line to y = xsin x at the point (π, 1). Taking logarithms we have ln y = sin x ln x, and dierentiating we nd 1 dy sin x = cos x ln x + . y dx x When x = π we have y = 1 and sin π = 0 so y 0 (π) = cos π ln π = − ln π . The tangent line is therefore y = 1 − (x − π) ln π . [3] c. We have z(t) = ex(t)·y(t) where x, y both depend on t. If at t = 1 we have x(1) = 0, x0 (1) = 1, y(1) = 2, y 0 (1) = 3 nd dz dt at t = 1. Dierentiating using the chain rule we have: dz d = ex(t)y(t) (xy) dt dt = exy (x0 y + y 0 x) . At t = 1 we have x(1) = 0 so z 0 (1) = e0 x0 (1)y(1) = 1 · 1 · 2 = 2 . Math 100 105, Fall 2010 Midterm, page 3/7 Problem 1 [3] d. The population of Canada was roughly 18 million in the year 1960, 30 million in the year 2000. Assuming the population grows exponentially, estimate the population of Canada in 2040. 5 In 40 years the population grew by a factor of about 30 18 = 3 . We thus expect the population in 2040 to be about 5 · 30million = 50million . 3 [3] e. Let f (x) = ex + e−x . Use a 2nd order Taylor polynomial to give a rational number approximating f ( 12 ). To second order we have ex ≈ 1 + x + we nd At x = 1 2 x2 2 so e−x ≈ 1 + (−x) + (−x)2 2 = 1−x+ x2 2 . Adding the two ex + e−x ≈ 2 + x2 . this reads 1 e1/2 + e−1/2 ≈ 2 . 4 [3] f. Show that the error in the approximation is less than that 2 ≤ e ≤ 3. 1 50 . You may use the fact We note that the expansion above is correct to 3rd order (the next term would be error is therefore at most (4) 4 f (c) 1 24 2 x3 6 3 + (−x) 6 ). The for some 0 ≤ c ≤ 12 . Now the fourth derivative of ex is ex , and of e−x is (−1)4 e−x = e−x so √ √ f (4) (x) = f (x) = ex + e−x . For 0 ≤ c ≤ 21 we then have f (4) (c) = ec + e−c ≤ e + e0 ≤ 4 + 1 = 3. It follows that the error is at most 1 1 1 3 1 · = = < . 24 16 8 · 16 128 50 Math 100 105, Fall 2010 2 Midterm, page 4/7 Problem 2 Long-form answers [6] Find the maximum value of f (x) = x q 1 − 34 x2 on the interval [0, 1]. q • For |x| < 1 we have 43 x2 ≤ 34 < 1 so by the chain rule 1 − 34 x2 is dierentiable in the open interval (0, 1). By the product rule the same holds for f . f is also continuous where dened, so it is continuous in the closed interval. It follow that the maximum of f occurs either at an endpoint or at a critical point. q • We have f (0) = 0 and f (1) = 1 − 43 = 12 . • We have f 0 (x) = q 1 − 43 x2 + x √ − 32 x = q f 0 (x0 ) = 0 when 32 x20 = 1, that is for x0 = 23 . 2 1 1− 34 x2 q q q • At the critical point, we have f ( 23 ) = 23 1 − • Finally, since 3 < 4 we have √ 3< √ 4 so √1 3 > 1 2 3 4 · 1− 34 x2 − 43 x2 √ 2 3 = 1− 43 x2 q 2 3 3 2 1− x = √ 2 3 2 .It follows that 1− 1− 4 x 1 2 = √1 . 3 > 0 so the absolute maximum is √1 . 3 Math 100 105, Fall 2010 3 Midterm, page 5/7 Problem 3 Long-form answers [6] A point is moving on the curve y2 = x3 − 3 in such a way that the x-co-ordinate . √How fast is the distance of the point to the origin is changing at the rate of 3 changing, when the point is at (2, 5)? units min The distance of the point from the origin is p p D(t) = x2 + y 2 = x2 + x3 − 3 . It follows that dD 1 d 2 = √ (x + x3 − 3) 2 3 dt dt 2 x +x −3 dx 2x + 3x2 = √ . 2 3 2 x + x − 3 dt When x = 2 and dx dt = 3 units this gives min dD 4 + 12 units = √ ·3 dt min 2 4+8−3 units 16 = √ ·3 min 2 9 units . =8 min Math 100 105, Fall 2010 4 Midterm, page 6/7 Problem 4 Long-form answers Let f (x) = ex − e−x . [1] Verify that (f 0 (x))2 = 4 + (f (x))2 . 2 We have f 0 (x) = ex + e−x . Since ex · e−x = ex−x = 1 we have (f (x)) = e2x − 2 + e−2x and 2 (f 0 (x)) = e2x + 2 + e−2x = 4 + e2x − 2 + e−2x . [3] Show that f has an inverse function. Three approaches: 2 1. From the rst part we know that (f 0 (x)) ≥ 4 and in particular it is never zero. Since f 0 (0) = 1 + 1 = 2 > 0 we nd that f 0 (0) > 0 everywhere (f 0 is continuous, so if it took negative and positive values it would vanish). It follows that f is strictly monotone and so has an inverse function. 2. We know that ex > 0 for all x. It follows that f 0 (x) = ex + e−x > 0 everywhere. 3. Let y be any real number. We will show that that is a unique x such that f (x) = y . Consider the equation t − 1t = y for an unkown t > 0 (t stands for ex ), equivalently t2 − yt − 1 = 0. √ q q y± y 2 +4 y y 2 y 2 + 1 . Since + 1 > y2 the solution This has the two solutions = ± 2 2 2 2 q q y y 2 y 2 y − + 1 is negative. On the other hand, the solution t = + + 1 is positive 2 2 2 2 1 (even if y is negative). So there is a unique positive value of t such that t + t = y . Given t, there is a unique x such that ex = t (that is x = ln t) so indeed for every y there is a unique x such that f (x) = y in other words, f has an inverse function. That function is " # r y 2 y x = ln + +1 . 2 2 Math 100 105, Fall 2010 [3] Let x, y . x = g(y) Midterm, page 7/7 Problem 4 be the inverse function. Find a formula for its derivative in terms of By the inverse function rule we have g 0 (y) = 1 f 0 (x) = ex 1 . + e−x If you took the third approach before and found g(y) = ln g (y) = y 2 = y 2 y 2 + q y 2 2 " 1 =p . y2 + 4 [3] Find a formula for g0 (y) involving y alone. From the rst part we know that f (x) = 0 + 1 , you would now nd # y/2 1 q + p 2 2 (y/2)2 + 1 y 2 + 1 + 2 "p # (y/2)2 + 1 + y/2 1 q p y 2 2 (y/2)2 + 1 +1 + 2 1 0 q 2 4 + (f (x)) = g 0 (y) = p 1 4 + y2 . p 4 + y 2 so